$\overline{AB} = 2\sqrt{41}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $2\sqrt{41}$ $?$ $ \sin( \angle BAC ) = \frac{5\sqrt{41} }{41}, \cos( \angle BAC ) = \frac{4\sqrt{41} }{41}, \tan( \angle BAC ) = \dfrac{5}{4}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{2\sqrt{41}} $ $ \overline{AC}=2\sqrt{41} \cdot \cos( \angle BAC ) = 2\sqrt{41} \cdot \frac{4\sqrt{41} }{41} = 8$